Arithmetic Sequences & Series ($u_n$ and $S_n$)
- Definition: An arithmetic sequence is an ordered progression where the difference between consecutive terms is constant: $$d = u_{n+1} - u_n = \text{constant (common difference)}$$
- General $n^{\text{th}}$ Term Formula: $$u_n = u_1 + (n - 1)d$$ where $u_1$ is the first term, $d$ is the common difference, and $n \in \mathbb{Z}^+$ is the term position.
- Sum of First $n$ Terms ($S_n$): Derived from Gauss's pairing method ($S_n = u_1 + u_2 + \dots + u_n$ added to its reverse): $$S_n = \frac{n}{2}(u_1 + u_n) = \frac{n}{2}\big(2u_1 + (n - 1)d\big)$$
Example 1 · Criterion A
Finding Unknowns in an Arithmetic Sequence
An arithmetic sequence has $5^{\text{th}}$ term $u_5 = 19$ and $11^{\text{th}}$ term $u_{11} = 43$.
(a) Find the first term $u_1$ and common difference $d$.
(b) Find the $30^{\text{th}}$ term $u_{30}$.
(c) Calculate the sum of the first $30$ terms $S_{30}$.
- (a) Setting up simultaneous equations: $$u_5 = u_1 + 4d = 19 \quad \text{--- (1)}$$ $$u_{11} = u_1 + 10d = 43 \quad \text{--- (2)}$$ Subtract (1) from (2): $$(u_1 + 10d) - (u_1 + 4d) = 43 - 19 \implies 6d = 24 \implies d = 4$$ Substitute $d = 4$ into (1): $$u_1 + 4(4) = 19 \implies u_1 + 16 = 19 \implies u_1 = 3$$
- (b) Finding $u_{30}$:: $$u_{30} = u_1 + (30 - 1)d = 3 + 29(4) = 3 + 116 = 119$$
- (c) Calculating $S_{30}$: $$S_{30} = \frac{30}{2}(u_1 + u_{30}) = 15(3 + 119) = 15(122) = 1830$$
Answers: (a) $u_1 = 3, d = 4$; (b) $u_{30} = 119$; (c) $S_{30} = 1830$.
Geometric Sequences & Finite Series ($u_n$ and $S_n$)
- Definition: A geometric sequence is an ordered progression where the ratio of consecutive terms is constant: $$r = \frac{u_{n+1}}{u_n} = \text{constant (common ratio)} \quad (r \ne 0)$$
- General $n^{\text{th}}$ Term Formula: $$u_n = u_1 \cdot r^{n-1}$$
- Finite Geometric Series Sum Formula ($S_n$): $$S_n = \frac{u_1(1 - r^n)}{1 - r} = \frac{u_1(r^n - 1)}{r - 1} \quad (r \ne 1)$$
- Infinite Geometric Series Preview (Extended): If the common ratio satisfies $|r| < 1$, then $r^n \to 0$ as $n \to \infty$, yielding a convergent sum to infinity: $$S_\infty = \frac{u_1}{1 - r} \quad (|r| < 1)$$
Example 2 · Criterion A
Geometric Progression Terms and Sum
In a geometric sequence, $u_1 = 12$ and $u_4 = 96$.
(a) Find the common ratio $r$.
(b) Find the $8^{\text{th}}$ term $u_8$.
(c) Calculate the sum of the first $8$ terms $S_8$.
- (a) Common ratio: $$u_4 = u_1 \cdot r^3 \implies 96 = 12 \cdot r^3$$ $$r^3 = \frac{96}{12} = 8 \implies r = \sqrt[3]{8} = 2$$
- (b) Finding $u_8$: $$u_8 = u_1 \cdot r^7 = 12 \cdot 2^7 = 12 \cdot 128 = 1536$$
- (c) Calculating $S_8$: $$S_8 = \frac{u_1(r^8 - 1)}{r - 1} = \frac{12(2^8 - 1)}{2 - 1} = \frac{12(256 - 1)}{1} = 12 \times 255 = 3060$$
Answers: (a) $r = 2$; (b) $u_8 = 1536$; (c) $S_8 = 3060$.
Investigating Patterns & Recursive Sequences (Criterion B)
- MYP Criterion B Inquiry Workflow:
1. Generate raw data from a geometric or numerical pattern.
2. Organize table of terms ($n$ vs $u_n$).
3. Compute first differences $\Delta u_n = u_{n+1} - u_n$ (constant $\implies$ linear/arithmetic, $u_n = an + b$).
4. Compute second differences $\Delta^2 u_n$ (constant $\implies$ quadratic, $u_n = an^2 + bn + c$ with $2a = \text{2nd diff}$).
5. Compute ratios $\frac{u_{n+1}}{u_n}$ (constant $\implies$ geometric, $u_n = u_1 \cdot r^{n-1}$).
6. State general rule in terms of $n$ and verify with subsequent terms. - Recursive Definitions:
A recurrence relation defines $u_{n+1}$ using preceding terms:
• Arithmetic: $u_{n+1} = u_n + d$, $u_1 = a$.
• Geometric: $u_{n+1} = r \cdot u_n$, $u_1 = a$.
• Fibonacci: $F_{n+1} = F_n + F_{n-1}$ with $F_1 = 1, F_2 = 1$.
For the arithmetic sequence $7, 12, 17, 22, \dots$:
(a) State the first term $u_1$ and common difference $d$. [1 mark]
(b) Write down a formula for the $n^{\text{th}}$ term $u_n$ in simplest form. [1 mark]
(c) Find the $25^{\text{th}}$ term. [1 mark]
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(a) $u_1 = 7$, $d = 12 - 7 = \mathbf{5}$ [A1]
(b) $u_n = 7 + (n - 1)5 = 7 + 5n - 5 = \mathbf{5n + 2}$ [A1]
(c) $u_{25} = 5(25) + 2 = 125 + 2 = \mathbf{127}$ [A1]
For the geometric sequence $3, -6, 12, -24, \dots$:
(a) State the common ratio $r$. [1 mark]
(b) Find the $10^{\text{th}}$ term $u_{10}$. [1 mark]
(c) Calculate the sum of the first $10$ terms $S_{10}$. [1 mark]
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(a) $r = \frac{-6}{3} = \mathbf{-2}$ [A1]
(b) $u_{10} = 3 \times (-2)^{10 - 1} = 3 \times (-2)^9 = 3 \times (-512) = \mathbf{-1536}$ [A1]
(c) $S_{10} = \frac{3(1 - (-2)^{10})}{1 - (-2)} = \frac{3(1 - 1024)}{3} = 1 - 1024 = \mathbf{-1023}$ [A1]
In an arithmetic sequence, $u_1 = 82$ and $d = -3.5$.
(a) Determine which term is the first negative term of the sequence. [2 marks]
(b) Find the value of this first negative term. [2 marks]
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(a) Finding position $n$: [2 marks]
Set $u_n < 0$: $82 + (n - 1)(-3.5) < 0$ [M1]
$82 - 3.5n + 3.5 < 0 \implies 85.5 < 3.5n$
$n > \frac{85.5}{3.5} \approx 24.428\dots$
Since $n$ must be an integer, $n = \mathbf{25}$ ($25^{\text{th}}\text{ term}$). [A1]
(b) Value: [2 marks]
$u_{25} = 82 + 24(-3.5) = 82 - 84 = \mathbf{-2}$ [A2]
Calculate the sum of all multiples of $4$ between $50$ and $250$ inclusive.
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Step 1: Identify first and last terms [1 mark]
First multiple $\ge 50$ is $u_1 = 52$ ($4 \times 13$) [A1]
Last multiple $\le 250$ is $u_n = 248$ ($4 \times 62$)
Step 2: Number of terms $n$ [1 mark]
$248 = 52 + (n - 1)4 \implies 196 = 4(n - 1) \implies n - 1 = 49 \implies n = 50$ [A1]
Step 3: Sum $S_{50}$ [2 marks]
$S_{50} = \frac{50}{2}(u_1 + u_{50}) = 25(52 + 248) = 25(300) = \mathbf{7500}$ [A2]
A sequence of matchstick patterns forms a series of squares in a row:
• Pattern 1 (1 square): 4 matchsticks
• Pattern 2 (2 squares joined): 7 matchsticks
• Pattern 3 (3 squares joined): 10 matchsticks
(a) Tabulate the matchstick counts for $n = 1, 2, 3, 4, 5$. [1 mark]
(b) Identify the sequence type and state a formula for $M_n$, the number of matchsticks in Pattern $n$. [2 marks]
(c) Determine the number of squares that can be formed using a box of $250$ matchsticks, and find the number of leftover matchsticks. [2 marks]
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(a) Table: [1 mark]
$n = 1 \to 4$; $n = 2 \to 7$; $n = 3 \to 10$; $n = 4 \to 13$; $n = 5 \to 16$. [A1]
(b) Formula: [2 marks]
Common difference is $d = +3$. Arithmetic sequence [M1].
$M_n = 4 + (n - 1)3 = \mathbf{3n + 1}$ [A1]
(c) Squares from 250 sticks: [2 marks]
$3n + 1 \le 250 \implies 3n \le 249 \implies n \le 83$.
$n = \mathbf{83\text{ squares}}$ [A1]
Matchsticks used: $3(83) + 1 = 250$. Leftover sticks = $\mathbf{0}$. [A1]
Prove the finite geometric series sum formula by algebra: $$S_n = \frac{u_1(1 - r^n)}{1 - r} \quad (r \ne 1)$$
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Step 1: Write series definition [1 mark]
$S_n = u_1 + u_1 r + u_1 r^2 + \dots + u_1 r^{n-1} \quad \text{--- (1)}$ [M1]
Step 2: Multiply by $r$ [1 mark]
$r S_n = u_1 r + u_1 r^2 + u_1 r^3 + \dots + u_1 r^n \quad \text{--- (2)}$ [M1]
Step 3: Subtract equations [1 mark]
$S_n - r S_n = u_1 - u_1 r^n$ [M1]
(Notice that all intermediate terms $u_1 r, u_1 r^2, \dots, u_1 r^{n-1}$ cancel in pairs).
Step 4: Factor out $S_n$ and divide [1 mark]
$S_n(1 - r) = u_1(1 - r^n)$
Since $r \ne 1$, divide by $(1 - r)$:
$S_n = \frac{u_1(1 - r^n)}{1 - r}$ $\blacksquare$ [R1]
The sum to infinity of a convergent geometric sequence is $S_\infty = 48$, and its first term is $u_1 = 36$.
(a) Find the common ratio $r$. [2 marks]
(b) Calculate the sum of the first $5$ terms $S_5$ as an exact fraction. [2 marks]
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(a) Common ratio: [2 marks]
$S_\infty = \frac{u_1}{1 - r} \implies 48 = \frac{36}{1 - r}$ [M1]
$1 - r = \frac{36}{48} = \frac{3}{4} \implies r = 1 - \frac{3}{4} = \mathbf{\frac{1}{4}}$ [A1]
(b) Sum $S_5$: [2 marks]
$S_5 = \frac{36(1 - (1/4)^5)}{1 - 1/4} = \frac{36(1 - 1/1024)}{3/4} = 36 \times \frac{4}{3} \times \frac{1023}{1024} = 48 \times \frac{1023}{1024}$
$S_5 = \frac{3 \times 1023}{64} = \mathbf{\frac{3069}{64}}$ ($= 47.953125$) [A2]
A superball is dropped from a height of $h_0 = 10\text{ metres}$. Each time it bounces off the floor, it rebounds to $80\%$ ($0.8$) of its preceding height.
(a) Find the height reached after the $4^{\text{th}}$ bounce. [2 marks]
(b) Determine on which bounce the rebound height first drops below $1\text{ metre}$. [2 marks]
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(a) Height on 4th bounce: [2 marks]
$h_n = 10 \times (0.8)^n$
$h_4 = 10 \times (0.8)^4 = 10 \times 0.4096 = \mathbf{4.096\text{ metres}}$ [A2]
(b) Bounce position: [2 marks]
$10 \times (0.8)^n < 1 \implies (0.8)^n < 0.1$
Take logarithms: $n \ln(0.8) < \ln(0.1)$
Since $\ln(0.8) \approx -0.2231 < 0$, reverse inequality [M1]:
$n > \frac{\ln 0.1}{\ln 0.8} = \frac{-2.3026}{-0.2231} \approx 10.32$
The rebound height drops below $1\text{ m}$ on the $\mathbf{11^{\text{th}}\text{ bounce}}$. [A1]
A student deposits $\$200$ at the start of each month into a savings account that pays $0.5\%$ interest per month compounded monthly.
• The $1^{\text{st}}$ deposit earns interest for $24$ months: $200(1.005)^{24}$.
• The $2^{\text{nd}}$ deposit earns interest for $23$ months: $200(1.005)^{23}$.
• The final deposit earns interest for $1$ month: $200(1.005)^1$.
(a) Identify the sequence type formed by the compounded monthly values and state $u_1$ and $r$. [2 marks]
(b) Use the geometric series sum formula to calculate the total balance in the account at the end of $2\text{ years}$ ($24\text{ months}$). [3 marks]
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(a) Identification: [2 marks]
Ordering from last deposit to first deposit:
$u_1 = 200(1.005) = 201$, $r = 1.005$, $n = 24$. Geometric series. [A2]
(b) Sum calculation: [3 marks]
$S_{24} = \frac{u_1(r^{24} - 1)}{r - 1} = \frac{201((1.005)^{24} - 1)}{1.005 - 1}$ [M1]
$(1.005)^{24} \approx 1.1271598$
$S_{24} = \frac{201(0.1271598)}{0.005} = 201 \times 25.431956 \approx \mathbf{\$5111.82}$ [A2]
Triangular numbers represent bowling pin or coin triangular stacks:
$T_1 = 1$, $T_2 = 1 + 2 = 3$, $T_3 = 1 + 2 + 3 = 6$, $T_4 = 1 + 2 + 3 + 4 = 10$.
(a) State the values of $T_5$ and $T_6$. [1 mark]
(b) Using the arithmetic sum formula, derive the closed formula $T_n = \frac{n(n+1)}{2}$. [2 marks]
(c) Investigate the sum of any two consecutive triangular numbers $T_n + T_{n-1}$. Prove algebraically that this sum is always a perfect square. [2 marks]
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(a) Next terms: [1 mark]
$T_5 = 10 + 5 = \mathbf{15}$, $T_6 = 15 + 6 = \mathbf{21}$ [A1]
(b) Derivation: [2 marks]
$T_n$ is an arithmetic series with $u_1 = 1, d = 1, u_n = n$:
$T_n = \frac{n}{2}(u_1 + u_n) = \frac{n}{2}(1 + n) = \mathbf{\frac{n(n+1)}{2}}$ [A2]
(c) Sum of consecutive triangular numbers: [2 marks]
$T_n + T_{n-1} = \frac{n(n+1)}{2} + \frac{(n-1)n}{2} = \frac{n^2 + n + n^2 - n}{2} = \frac{2n^2}{2} = \mathbf{n^2}$
Since $n^2$ is the square of integer $n$, the sum is always a perfect square. $\blacksquare$ [R2]
Sequences & Series Quick Reference
Arithmetic Progression (AP)
Geometric Progression (GP)
Infinite Convergent GP (Extended)
Triangular & Special Numbers
Top 4 Sequences Pitfalls
1. The "$n$" vs "$n - 1$" Trap
In geometric sequences, students often write $u_n = u_1 r^n$. This is wrong! The first term has 0 multiplications by $r$, so the exponent is strictly $n - 1$: $u_n = u_1 r^{n-1}$.
2. Using $S_\infty$ When $|r| \ge 1$
The formula $S_\infty = \frac{u_1}{1 - r}$ ONLY applies if the terms shrink toward zero ($|r| < 1$). If $r = 2$ or $r = -1.5$, the infinite series diverges.
3. Confusing Sequence ($u_n$) with Series ($S_n$)
$u_n$ is the value of the single $n^{\text{th}}$ term. $S_n$ is the total sum of all terms from $1$ to $n$. When asked for total money saved or distance traveled, use $S_n$, not $u_n$.
4. Signs in Alternating GP Series
When $r < 0$ (e.g. $r = -2$), always put brackets around $r$: $1 - (-2)^{10} = 1 - 1024 = -1023$, and $1 - (-2) = 1 + 2 = 3$.
Number Sequences Check
Answer these 5 questions to test AP, GP, series sums, and pattern reasoning.
An arithmetic sequence has $u_1 = 14$ and $d = -3$. What is the value of the $20^{\text{th}}$ term $u_{20}$?
What is the sum of the first $15$ terms of the arithmetic series $2 + 5 + 8 + 11 + \dots$?
In a geometric sequence, $u_1 = 5$ and common ratio $r = 3$. What is the sum of the first $6$ terms $S_6$?
For which value of common ratio $r$ does a geometric sequence converge to a finite sum to infinity $S_\infty$?
A theatre has $20$ seats in row 1, $24$ seats in row 2, $28$ seats in row 3, and continues this pattern for $15$ rows. What is the total seating capacity of the theatre?